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I Why are some systems defined with the name "variable mass"?

  1. Apr 8, 2016 #1
    I do not get why systems such as the rocket in space are defined as "variable mass" since the mass of the system is not varying.
    The equation used for such systems $$\sum F^{(E)}=\frac{d\vec{P}}{dt} \tag{1}$$ (sum of external forces on the system equals the change in momentum) holds true only if the total mass of the system does not change during the time interval [itex]dt[/itex] considered. Is this correct? I'm not sure of what I'm saying, only supposing it, because in the proof for [itex](1)[/itex] on textbook the center of mass is used and the mass of the system is taken as constant in the derivatives.
    In rocket motion for istance we consider a time interval [itex]dt[/itex] in which the mass of the rocket(with the gas inside of it) decreases of a quantity d[itex]m[/itex], but that mass of gas [itex]dm[/itex] is still in the system (rocket+ mass [itex]dm[/itex] of gas), even if it is not in the rocket anymore. In fact, writing the momentum of the system we do include the mass [itex]dm[/itex].
    $$P(t)=mv$$
    $$P(t+dt)=(m-dm)(v+dv)+dm(v-u)$$
    (Where [itex]u[/itex] is the relative velocity of the gas)
    Does the total mass of the system really increase or decrease and [itex](1)[/itex] holds true also if the total mass of the system is varying?
     
  2. jcsd
  3. Apr 8, 2016 #2
    The mass of a rocket varies due to the ejection of reaction mass.

    This equation also holds if the total mass changes.

    That depends on the definition of the system boundaries. It doesn't make much sense to include ejected mass into the system which doesn't interact with the rocket anymore.
     
  4. Apr 8, 2016 #3
    If you have two masses ##m_1## and ##m_2##, connected together, traveling as a single object at velocity ##v##, and then suddenly consider ##m_2## to no longer be part of the object, then the momentum of the "object" decreases from ##(m_1+m_2)v## to ##m_1 v##.
    There isn't any force involved with this change in momentum because it isn't a "real" change in momentum but a change in your definition of your object.

    If you have a closed system, the total momentum is conserved, so if you have an "object" which loses mass, you can easily calculate the change in momentum of the object if you know the rate that momentum and mass are ejected from the object.
     
  5. Apr 9, 2016 #4

    Thanks for the answer @DrStupid ! I think that it does not make sense to include the gas too, but when I write the equation [itex]P(t+dt)=(m-dm)(v+dv)+dm(v-u)[/itex] I do include the infinitesimal mass [itex] dm [/itex] of the gas that has been ejected. So the system considered is rocket+gas.
    Nevertheless this is quite confusing because it seems to be the system "just for" the time interval considered [itex]dt[/itex]. I'll try to explain it.
    The system at [itex]t[/itex] is rocket and gas (total mass [itex] m[/itex]), at [itex]t+dt[/itex] is the rocket and remaing gas (mass [itex]m-dm[/itex]) plus the ejected gas [itex]dm[/itex]. But after that? At the "next time interval" the mass of the rocket (with the gas inside of it) is still decreased of another amount [itex]dm[/itex] and the mass of the gas ejected is increased of the same amount. But in the equation only [itex] dm [/itex] appears, so it looks like we do not consider in the system the mass of the gas that has been ejected before that moment, but only the infinitesimal mass ejected in the time interval considered.

    Probably I misunderstood something here, would you be so kind as to give me any suggestions about it?


    Thanks for the reply @Khashishi! I got what you said but the "object" you mentioned is only a part of the system, right? (In the case of the rocket , the object is the rocket, which is loosing mass, but the system, rocket + gas, is not loosing mass).
     
  6. Apr 10, 2016 #5
    I don't think that this is correct. It should be

    [itex]p\left( {t + dt} \right) = m\left( {t + dt} \right) \cdot v\left( {t + dt} \right) = \left( {m - dm} \right) \cdot \left( {v + dv} \right)[/itex]

    I would avoid finite difference equation and use differential equations instead. With the second law

    [itex]F = \dot p[/itex]

    the forces acting on rocket and gas are

    [itex]F_R = m_R \cdot \dot v_R + v_R \cdot \dot m_R[/itex]
    [itex]F_G = m_G \cdot \dot v_G + v_G \cdot \dot m_G[/itex]

    With the third law

    [itex]F_R + F_G = 0[/itex]

    conservation of mass

    [itex]\dot m_R + \dot m_G = 0[/itex]

    mass ejection at constant velocity

    [itex]\dot v_G = 0[/itex]

    and the velocity of the ejected gas

    [itex]v_G = v_R + u[/itex]

    this turns into the rocket equation

    [itex]\dot v_R = u \cdot \frac{{\dot m_R }}{{m_R }}[/itex]
     
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